158
Electric Motors and Drives
produces the full rated torque of the motor. If now the load torque on the motor shaft is reduced to half rated torque, and assuming that the resistance is negligible, the steady-state speed will remain the same but the new mean steady-state current will be halved, as shown by the lower dotted curve. We note however that although, as expected, the mean current is determined by the load, the ripple current is unchanged, and this is explained below. If we ignore resistance, the equation governing the current during the ‘on’ period is V ¼EþL
di , dt
di 1 ¼ (V $ E) dt L
or
(4:4)
Since V is greater than E, the gradient of the current (di/dt) is positive, as can be seen in Figure 4.13(c). During this ‘on’ period the battery is supplying power to the motor. Some of the energy is converted to mechanical output power, but some is also stored in the magnetic Weld associated with the inductance. The latter is given by 1=2Li2 , and so as the current (i) rises, more energy is stored. During the ‘oV ’ period, the equation governing the current is 0¼EþL
di , dt
or
di E ¼$ dt L
(4:5)
We note that during the ‘oV ’ time the gradient of the current is negative (as shown in Figure 4.13(c)) and it is determined by the motional e.m.f. E. During this period, the motor is producing mechanical output power which is supplied from the energy stored in the inductance; not surprisingly the current falls as the energy previously stored in the ‘on’ period is now given up. We note that the rise and fall of the current (i.e. the current ripple) is inversely proportional to the inductance, but is independent of the mean d.c. current, i.e. the ripple does not depend on the load. To study the input/output power relationship, we note that the battery current only Xows during the ‘on’ period, and its average value is therefore kIdc . Since the battery voltage is constant, the power supplied is simply given by V (kIdc ) ¼ kVIdc . Looking at the motor side, the average voltage is given by Vdc ¼ kV , and the average current (assumed constant) is Idc , so the power input to the motor is again kVIdc , i.e. there is no loss of power in the ideal chopper. Given that k is less than one, we see that the input (battery) voltage is higher than the output (motor) voltage, but conversely the input current is less than the output current, and in this respect we see that the chopper